What is the center of the circle and the radius of 2x^2 - 12x +2y^2 +8y = 10

the center of the circle, (h, k), is (3, -2),

and the radius is √18, or 3√2.  

Step-by-step explanation:

This problem becomes much easier if you divide it through by 2:

x^2 - 6x +y^2 +4y = 5

Next, regroup this equation so that the x terms are together and the y terms are also together, separately:

x^2 - 6x + y^2 + 4y = 5

Next, complete the square of x^2 - 6x:

x^2 - 6x → x^2 - 6x + 9 - 9 →  (x - 3)^2 - 9.

Then complete the square of y^2 + 4y:

y^2 + 4y + 4  -  4  →  (y + 2)^2 - 4

Now rewrite x^2 - 6x + y^2 + 4y = 5 as

                       (x - 3)^2 - 9 + (y + 2)^2 - 4 = 5

Next, group the constants together on the right side:

     (x - 3)^2 - 9 + (y + 2)^2 - 4 = 5    →   (x - 3)^2 + (y + 2)^2  = 18  

Comparing this result to the standard equation of a circle:

                                                        (x - h)^2 + (y - k)^2 = r^2,

we see that h = 3, y = -2 and r^2 = 18.

Thus, the center of the circle, (h, k), is (3, -2),

and the radius is √18, or 3√2.        

55? i think maybe

step-by-step explanation: